The image contains the following mathematical problem:

Evaluate the limit:

$\lim_{{x \to 2}} \frac{{x^3 - 8}}{{x^2 - 4}}$

To solve this limit, we first notice that substituting $x = 2$ directly into the expression would yield an indeterminate form of $\frac{0}{0}$. Therefore, we need to simplify the expression before evaluating the limit.

Step 1: Factor the numerator and the denominator

• The numerator $x^3 - 8$ is a difference of cubes, which can be factored as: $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$
• The denominator $x^2 - 4$ is a difference of squares, which can be factored as: $x^2 - 4 = (x - 2)(x + 2)$

Step 2: Simplify the expression

The expression now becomes: $\frac{{(x - 2)(x^2 + 2x + 4)}}{{(x - 2)(x + 2)}}$ We can cancel the common factor $(x - 2)$ in both the numerator and the denominator: $\lim_{{x \to 2}} \frac{{x^2 + 2x + 4}}{{x + 2}}$

Step 3: Evaluate the limit

Now, substitute $x = 2$ into the simplified expression: $\frac{{2^2 + 2(2) + 4}}{{2 + 2}} = \frac{{4 + 4 + 4}}{{4}} = \frac{{12}}{{4}} = 3$

So, the limit is: $\boxed{3}$

Would you like any further details or explanations?

Related Questions:

1. How do you factor polynomials involving cubes?
2. What are common indeterminate forms in limits, and how do you resolve them?
3. Can you explain how L'Hopital's rule works for resolving indeterminate forms?
4. How do you approach limits that result in infinity?
5. What is the significance of simplifying expressions before evaluating limits?

Tip: Always check if you can factor the expression before directly substituting the value when evaluating limits. It often simplifies the problem and avoids indeterminate forms.